PHP如何显示用户数据

Question

我正在尝试显示登录的唯一用户，然后将其存储到我的数据库中的表中，例如，如果我以 James 身份登录，我想显示数据库中 James 的数据并将他的 id 存储到另一个table.据我了解 $_SESSION['login_user'];应该返回唯一的用户用户名。

当我运行时，我得到这个错误，这是输出：

注意：未定义索引：C:\xampp\htdocs\bid.php 第 3 行中的 login_user

注意：未定义变量：第 8 行 C:\xampp\htdocs\bid.php 中的 SESSION 感谢您联系我们

登录.php

<?php
   include("config.php");
   session_start();

   if($_SERVER["REQUEST_METHOD"] == "POST") {
      // username and password sent from form 

      $myusername = mysqli_real_escape_string($conn,$_POST['username']);
      $mypassword = mysqli_real_escape_string($conn,$_POST['password']); 

      $sql = "SELECT customer_id FROM customer WHERE email_adress = '$myusername' and password = '$mypassword'";
      $_SESSION['username'] = $myusername
      $result = mysqli_query($conn,$sql);
      $row = mysqli_fetch_array($result,MYSQLI_ASSOC);
      $active = $row['customer_id'];

      $count = mysqli_num_rows($result);

      // If result matched $myusername and $mypassword, table row must be 1 row

 if($count >= 1) {
     $_SESSION['login_user'] = $myusername;

     header("location: index2.php");
  }else {
     $error = "Your Login Name or Password is invalid";
  }
   }
?>

bid.php：

<?php
    session_start();
    echo $_SESSION['username'];
require 'config.php';
//include "job.php" 
$jobid    = $_POST['job_id'];
$bid    = $_POST['bid'];
bidderid = $SESSION['username']
echo "$jobid";






$query   = "INSERT into bid (bid_amount,job_id) VALUES('" . $bid . "','" . $jobid . "')";
$success = $conn->query($query);

if (!$success) {
    die("Couldn't enter data: ".$conn->error);

}

echo "Thank You For Contacting Us <br>";

$conn->close();

 ?>

session.php

<?php
   include('config.php');
   session_start();

   $user_check = $_SESSION['login_user'];

   $ses_sql = mysqli_query($conn,"select email_adress from customer where email_adress = '$user_check' ");

   $row = mysqli_fetch_array($ses_sql,MYSQLI_ASSOC);

   $login_session = $row['email_adress'];

   if(!isset($_SESSION['login_user'])){
      header("location:login.php");
   }
?>

Answer 1

登录.php

<?php
   session_start();
   // session should start here

   include("config.php");


   if($_SERVER["REQUEST_METHOD"] == "POST") {
      // username and password sent from form 

      $myusername = mysqli_real_escape_string($conn,$_POST['username']);
      $mypassword = mysqli_real_escape_string($conn,$_POST['password']); 

      $sql = "SELECT customer_id FROM customer WHERE email_adress = '$myusername' and password = '$mypassword'";

      $_SESSION['username'] = $myusername; // you missed a semi colon

      $result = mysqli_query($conn,$sql);
      $row = mysqli_fetch_array($result,MYSQLI_ASSOC);
      $active = $row['customer_id'];

      $count = mysqli_num_rows($result);

      // If result matched $myusername and $mypassword, table row must be 1 row

 if($count >= 1) {

     $_SESSION['login_user'] = $myusername; // $myusername contains username of this user

     $_SESSION['login_data'] = $row; // $row contains user record from the db. including customer_id etc..


     header("location: index2.php");
  }else {
     $error = "Your Login Name or Password is invalid";
  }
   }
?>

现在，如果成功，$_SESSION['login_user'] 将在 bid.php 和所有其他页面中可用。

1. 在 session.php 中，请恢复到这些。

     <?php
      session_start(); 
      // this should be called first before including external files 
      include('config.php');
   
      $user_check = $_SESSION['login_user'];
   
      $ses_sql = mysqli_query($conn,"select email_adress from customer where email_adress = '$user_check' ");
   
      $row = mysqli_fetch_array($ses_sql,MYSQLI_ASSOC);
   
      $login_session = $row['email_adress'];
   
      if(!isset($_SESSION['login_user'])){
         header("location:login.php");
      }
   ?>
   

2. 在bid.php中请参考这个。

        $bidderid = $_SESSION['login_user']->username or  $_SESSION['login_user']['username']
   

希望这会有所帮助.. 编码愉快